Could you please tell me how to iterate over list where items can contain whitespaces?
x=("some word", "other word", "third word")
for word in $x ; do
echo -e "$word\n"
done
how to force it to output:
some word
other word
third word
instead of:
some
word
(...)
third
word
To loop through items properly you need to use ${var[@]}. And you need to quote it to make sure that the items with spaces are not split: "${var[@]}".
All together:
x=("some word" "other word" "third word")
for word in "${x[@]}" ; do
echo -e "$word\n"
done
Or, saner (thanks Charles Duffy) with printf:
x=("some word" "other word" "third word")
for word in "${x[@]}" ; do
printf '%s\n\n' "$word"
done
set -f to turn globbing off. Please see wiki.bash-hackers.org/commands/builtin/set
echo -e? If what the OP really wants is an extra newline on the end, printf '%s\n\n' would be saner.Two possible solutions, one similar to fedorqui's solution, without the extra ',', and another using array indexing:
x=( 'some word' 'other word' 'third word')
# Use array indexing
let len=${#x[@]}-1
for i in $(seq 0 $len); do
echo -e "${x[i]}"
done
# Use array expansion
for word in "${x[@]}" ; do
echo -e "$word"
done
Output:
some word
other word
third word
some word
other word
third word
edit: fixed issues with the indexed solution as pointed out by cravoori
seq inside a command substitution. Also, start with 0 instead of 1